L-7 ECE-495 595
-
Upload
nathan-henry -
Category
Documents
-
view
24 -
download
0
Transcript of L-7 ECE-495 595
ECE 495/595 (aka ECE-381)
Introduction to Power Systems
Edward D. Graham, Jr. Ph.D., P.E.
Lecture 7September 7, 2011
Office: ECE-235B
More Three Phase Power
= instantaneous power
t
- Θi
Remember: Complex Conjugate (a + jb)* = (a – jb)
- Θi
0
NO
4Ω
0
Root
Square
Mean
2
PF = Power Factor
= Complex Power
This equation is the basis for the Power Triangle
Changing Landscape of Power Systems and Utilities Deregulation
Three Phase ExampleAssume a -connected load is supplied from a 313.8 kV (L-L) source with Z = 10020
13.8 013.8 013.8 0
ab
bc
ca
V kVV kVV kV
13.8 0 138 20
138 140 138 0
ab
bc ca
kVI amps
I amps I amps
Three Phase Example (continued)
*
138 20 138 0239 50 amps239 170 amps 239 0 amps
3 3 13.8 0 kV 138 amps5.7 MVA5.37 1.95 MVA
pf cos 20 lagging
a ab ca
b c
ab ab
I I I
I I
S V I
j
In the News: New CWLP GeneratorCWLP = City Water, Light & Power Springfield, Ill.
This is a 280 MVA Generator for CWLP’s New Coal Plant
Delta-Wye Transformation
Y
phase
To simplify analysis of balanced 3 systems:1) Δ-connected loads can be replaced by
1Y-connected loads with Z3
2) Δ-connected sources can be replaced byY-connected sources with V
3 30Line
Z
V
Delta-Wye Transformation Proof
From the side we get
Hence
ab ca ab caa
ab ca
a
V V V VIZ Z Z
V VZI
Delta Wye Transformation
Delta-Wye Transformation, (Continued)
a
From the side we get( ) ( )
(2 )Since I 0Hence 3
3
1Therefore3
ab Y a b ca Y c a
ab ca Y a b c
b c a b c
ab ca Y a
ab caY
a
Y
YV Z I I V Z I IV V Z I I I
I I I I IV V Z I
V VZ ZI
Z Z
Three Phase Transmission Line
Interconnected North American Power Grid
Three Phase Transmission Line
Per Phase Analysis
Per phase analysis allows analysis of balanced 3 systems with the same effort as for a single phase systemBalanced 3 Theorem: For a balanced
3 system with– All loads and sources Y connected– No mutual Inductance between phases
Per Phase Analysis (continued)
Then– All neutrals are at the same potential– All phases are COMPLETELY decoupled– All system values are the same sequence
as sources. The sequence order we’ve been using (phase b lags phase a and phase c lags phase a) is known as “positive” sequence (abc); later in the course we’ll discuss negative and zero sequence systems.
Per Phase Analysis ProcedureTo do per phase analysis1. Convert all load/sources to equivalent Y’s2. Solve phase “a” independent of the other
phases3. Total system power S = 3 Va Ia
*
4. If desired, phase “b” and “c” values can be determined by inspection (i.e., ±120° degree phase shifts)
5. If necessary, go back to original circuit to determine line-line values or internal values.
Per Phase ExampleAssume a 3, Y-connected generator with Van =
10 volts supplies a -connected load with Z = -j through a transmission line with impedance of j0.1 per phase. The load is also connected to a -connected generator with Va”b” = 10through a second transmission line which also has an impedance of j0.1 per phase.
Find1. The load voltage Va’b’2. The total power supplied by each generator, SY
and S
Per Phase Example (continued)
First convert the delta load and source to equivalent Y values and draw just the "a" phase circuit
Per Phase Example (continued)
' ' 'a a a
To solve the circuit, write the KCL equation at a'1(V 1 0)( 10 ) V (3 ) (V j3
j j
Per Phase Example (continued)
' ' 'a a a
'a
' 'a b' 'c ab
To solve the circuit, write the KCL equation at a'1(V 1 0)( 10 ) V (3 ) (V j3
10(10 60 ) V (10 3 10 )3
V 0.9 volts V 0.9 volts
V 0.9 volts V 1.56
j j
j j j j
volts
Per Phase Example (continued)
*'*
ygen
*" '"
S 3 3 5.1 3.5 VA0.1
3 5.1 4.7 VA0.1
a aa a a
a agen a
V VV I V jj
V VS V jj